Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 12A Problems/Problem 15"

(→‎Solution 5 (Easiest and best using shoelace))
(→‎Solution 5)
Line 95: Line 95:
 
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and  <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.
 
Without loss of generality, let the radius of the circle be equal to <math>3</math>. Thus, <math>AO=3</math> and <math>OB=3</math>. As a consequence of <math>2(AC)=BC</math>, <math>AC=2</math> and <math>CO=1</math>. Also, we know that <math>DO</math> and <math>OE</math> are both equal to <math>3</math> due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to <math>\sqrt{3^2-1^2}</math> or <math>2\sqrt{2}</math>. Now we know that the area of <math>[ABD]</math> is equal to <math>\frac{(3+2+1)(2\sqrt{2})}{2}</math> or <math>6\sqrt{2}</math>. Know we need to find the area of <math>[DCE]</math>. By simple inspection <math>[COD]</math> <math>\cong</math> <math>[HOE]</math> due to angles being equal and CPCTC. Thus <math>HE=2\sqrt{2}</math> and <math>OH=1</math>. Know we know the area of <math>[CHE]=\frac{(1+1)(2\sqrt{2})}{2}</math> or <math>2\sqrt{2}</math>. We also know that the area of <math>[OHE]=\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus the area of <math>[COE]=2\sqrt{2}-\sqrt{2}</math> or <math>\sqrt{2}</math>. We also can calculate the area of <math>[DOC]</math> to be <math>\frac{(1)(2\sqrt{2})}{2}</math> or <math>\sqrt{2}</math>. Thus <math>[DCE]</math> is equal to <math>[COE]</math> + <math>[DOC] </math> or <math>\sqrt{2}+\sqrt{2}</math> or <math>2\sqrt{2}</math>. The ratio between <math>[DCE]</math> and  <math>[ABD]</math> is equal to <math>\frac{2\sqrt{2}}{6\sqrt{2}}</math> or <math>\frac{1}{3}</math> <math>\Longrightarrow \mathrm{(C)}</math>.
 
===Solution 5===
 
===Solution 5===
−
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A,B, and C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.
+
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for <math>D (x,y)</math>, and notice how <math>E</math> is a 180 degree rotation of <math>D</math>, using the rotation matrix formula we get <math>E = (-x,-y)</math>. WLOG say that this circle has radius <math>3</math>. We can now find points <math>A</math>, <math>B</math>, and <math>C</math> which are <math>(-3,0)</math>, <math>(-1,0)</math>, and <math>(3,0)</math> respectively.
 
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.
 
By shoelace the area of <math>CED</math> is <math>Y</math>, and the area of <math>ADB</math> is <math>3Y</math>. Using division we get that the answer is <math> \mathrm{(C)}</math>.
  

Revision as of 08:22, 20 October 2019

Problem

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?

[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2));[/asy]

$(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}$

Solution

Solution 1

Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or $\frac{CF}{CD}$ ($F$ is the foot of the perpendicular from $C$ to $DE$).

Call the radius $r$. Then $AC = \frac 13(2r) = \frac 23r$, $CO = \frac 13r$. Using the Pythagorean Theorem in $\triangle OCD$, we get $\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r$.

Now we have to find $CF$. Notice $\triangle OCD \sim \triangle OFC$, so we can write the proportion:

$\frac{OF}{OC} = \frac{OC}{OD}$
$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$
$OF = \frac 19r$

By the Pythagorean Theorem in $\triangle OFC$, we have $\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r$.

Our answer is $\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}$.

Solution 2

Let the center of the circle be $O$.

Note that $2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB$.

$O$ is midpoint of $AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB$.

$O$ is midpoint of $DE \Rightarrow$ Area of $\triangle DCE = 2 \cdot$ Area of $\triangle DCO = 2 \cdot (\frac{1}{6} \cdot$ Area of $\triangle ABD) = \frac{1}{3} \cdot$ Area of $\triangle ABD \Longrightarrow \mathrm{(C)}$.

Solution 3

Let $r$ be the radius of the circle. Note that $AC+BC = 2r$ so $AC = \frac{2}{3}r$.

By Power of a Point Theorem, $CD^2= AC \cdot BC = 2\cdot AC^2$, and thus $CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r$

Then the area of $\triangle ABD$ is $\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2$. Similarly, the area of $\triangle DCE$ is $\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2$, so the desired ratio is $\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}$

Solution 4

[asy] unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), e=(-D.x,-D.y); pair H=(e.x,0); draw(A--B--D--cycle); draw(D--e--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",e,SSE); label("$B$",B,NE); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("$H$",H,SE); draw(e--H,dashed); label("O",(0,0),NE); label("1",(C--O),N); label("1",(H--O),N); label("2",(A--C),N); label("2",(H--B),N); label("3",(O--D),NE); label("3",(O--e),NE); label("$2\sqrt{2}$",(D--C),W); label("$2\sqrt{2}$",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));[/asy]

Let the center of the circle be $O$. Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\sqrt{3^2-1^2}$ or $2\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\frac{(3+2+1)(2\sqrt{2})}{2}$ or $6\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\frac{(1+1)(2\sqrt{2})}{2}$ or $2\sqrt{2}$. We also know that the area of $[OHE]=\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus the area of $[COE]=2\sqrt{2}-\sqrt{2}$ or $\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\sqrt{2}+\sqrt{2}$ or $2\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\frac{2\sqrt{2}}{6\sqrt{2}}$ or $\frac{1}{3}$ $\Longrightarrow \mathrm{(C)}$.

Solution 5

We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a 180 degree rotation of $D$, using the rotation matrix formula we get $E = (-x,-y)$. WLOG say that this circle has radius $3$. We can now find points $A$, $B$, and $C$ which are $(-3,0)$, $(-1,0)$, and $(3,0)$ respectively. By shoelace the area of $CED$ is $Y$, and the area of $ADB$ is $3Y$. Using division we get that the answer is $\mathrm{(C)}$.

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_15&oldid=110466"